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3 years ago

11

What is the interquartile range

1 answer:

motikmotik3 years ago

8 0

<em>Answer:</em>

<em></em>

<em>Step-by-step explanation:</em>

<em>Khan academy huh?</em>

<em>put the numbers in order from least to greatest.</em>
<em>find the median</em>
<em>find the difference between the median and all of the other numbers.</em>

<em></em>

<em></em>

<em>P.S. If you were looking for somebody to tell you the answer don't hold your breath.</em>

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Can anyone help thanks

Soloha48 [4]

Answer:

x = 8, -4

Step-by-step explanation:

2x-4-6=6; therefore, x = 8

2x-4=-12; therefore, x = -4

5 0

3 years ago

−3z−z<br><br><br><br> Help pleease i will give u 20 points

MAVERICK [17]

Answer:

-4z is the answer

Step-by-step explanation:

-3z-z

-4z

3 0

2 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

8 0

3 years ago

NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!

Georgia [21]

The third table on the top row. It has a linear function because the Y is always constant. For example: -3 + -2= -5 -5 + -2= -7

7 0

3 years ago

What is the sixth term of (x + y)8?<br> 56x3y5<br><br> 36x3y5<br><br> 36x4y4<br><br> 56x4y4

Brut [27]

It's either the first or second answer

4 0

3 years ago